Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 71 / 7

Model the polymerization Brownian ratchet by a rigid tip and a Brownian load. Let $a$ be the elementary load advance, ${\zeta }_L$ its drag coefficient, and $D=k_{B}T/{\zeta }_L$ its diffusion coefficient. For a spherical load, ${\zeta }_L=6\pi \mu R_L$, but no load radius is specified, so the rate is naturally expressed in terms of $D$. Assume addition is effectively instantaneous once the gap $h$ reaches $a$, and removal is negligible. The gap reflects at zero, first arrival at $a$ causes insertion, and insertion resets it to zero.

With no opposing force, the mean first-passage time satisfies $D{T}^{\prime \prime }=-1$, $T^{\prime }(0)=0$, $T(a)=0$. Integration gives $T(h)=(a^2-h^2)/(2D)$. One advance per passage givesThe advance-event rate is $2D/a^2$, while $V_0$ is a length per time.

A constant resisting force $F\ge 0$ produces drift $-u_F=-DF/(k_{B}T)$. The diffusion-limited Brownian ratchet with constant load now satisfiesAn integrating factor gives $T^{\prime }(h)=[1-e^{u_{F}h/D}]/u_F$. Integrating again from $h$ to $a$ gives $T(0)=D(e^z-1-z)/u_F^2$, where $z=Fa/(k_{B}T)$. ThereforeAt weak load the ratio is $1-z/3+O(z^2)$; at large load $V\sim (D/a)z^2{e}^{-z}$. Replacing the whole force dependence by $e^{-z}$ would discard the first-passage kinetics of this regime.

For two staggered strands that alternately advance the leading tip by half a monomer, $a=\delta /2$ and $V_0=4D/\delta$. Two aligned strands, as schematically drawn, do not automatically have that half-step: an advance may be $a=\delta$, with the other strand filling the same layer afterward. Use the same geometrical step in the passage problem and in converting event rate to velocity. The limiting formula assumes negligible removal, rigid tips, an overdamped load with constant drag, and fast insertion compared with $a^2/D$. With appreciable removal or additional internal strand states, the renewal reset model must be replaced by a coupled gap/reaction process.

Write $u$ for the effective unloaded addition rate and $w$ for the removal rate of the same elementary tip step $a$. If the supplied on-rate is a bimolecular rate constant, then $u=k_{\mathrm{on}}{c}_m$, with monomer concentration $c_m$; if it already denotes an event rate, use $u=k_{\mathrm{on}}$. Assume gap diffusion is much faster than chemical events, removal is force-independent, and the monomer bath is maintained.

Under a positive constant resisting force, the load equilibrates in the potential $Fh$ above a reflecting tip. The normalized Boltzmann distribution is $p(h)=\beta F{e}^{-\beta Fh}$ for $h\ge 0$, where $\beta =1/(k_{B}T)$. The fraction of time with room to insert isSteric gating reduces addition to $u{e}^{-\beta Fa}$; each addition advances by $a$ and each removal retreats by $a$. The reaction-limited polymerization Brownian ratchet therefore hasFor $u>w>0$, the stall force is $F_s=(k_{B}T/a)\log (u/w)$. Equivalently, the available chemical free energy per effective step is $\Delta \mu =k_{B}T\log (u/w)$, and stall balances $F_{s}a=\Delta \mu$. A half-monomer step requires rates defined for that half-step, not automatically single-strand monomer rates with a gratuitous factor of two.

At exactly zero force a load on an unbounded half-line has no normalizable equilibrium gap density. The unloaded expression is the $F\downarrow 0$ limit, or the usual freely available-gap approximation with a weak positional confinement. For a positive load the equilibration time must remain short compared with reaction times. Load inertia, filament bending, monomer depletion, and force-dependent bond rupture have been excluded. For arbitrary finite reaction and diffusion rates one must solve the joint gap dynamics rather than attach $-aw$ to the diffusion-limited first-passage formula: a removal changes the gap and the distribution from which the next addition starts.